Question

A biased coin is tossed twice. The probability of head is twice the tail. The PDF of the number of heads is

$x$	0	1	2
$P\left( x \right)$	$\dfrac{a}{d}$	$\dfrac{b}{d}$	$\dfrac{c}{d}$

Then the values of a, b, c, d are
A. a = 1, b = 2, c = 3, d = 4
B. a = 1, b = 4, c = 4, d = 9
C. a = 1, b = 4, c = 4, d = 10
D. a = 1, b = 2, c = 1, d = 4

Answer 1

Hint : To solve the given question, we need to know what a binomial distribution is. A binomial distribution can be thought of as simply the probability of a success or a failure outcome in an experiment. If we have n as the number of trials, x as the number of successes desired, p as the probability of getting a success in one trial and q as the probability of getting a failure in one trial, which is always 1 -p, then the probability distribution, $P\left( x \right)=n{{C}_{x}}{{p}^{x}}{{q}^{n-x}}$ . We will use this concept to get our desired answer.

Complete step-by-step answer :
We have been given the question that a biased coin is tossed twice and the probability of head is twice the tail. We have been given a PDF of the number of heads and have been asked to find the values of a, b, c, d. Here, we are given the experiment as tossing a biased coin twice, so the number of trials, n = 2 times. Now, let us consider the event of getting head as a success. Therefore, we get,
The probability of getting head in a trial = $p$ , and the probability of getting tail in a trial = $q$ .
We are also given that the probability of getting a head is twice that of the tail, so it means, $p=2q$ or $q=\dfrac{p}{2}$ . But we know that the total probability should be equal to 1. We have a binomial distribution here with two events, getting a head or a tail. So, the total probability will be,
$p+q=1$
We know that $p=2q$ , so we will substitute it in the above equation and get,
$2q+q=1\Rightarrow q=\dfrac{1}{3}$
So, it means, we can write the value of $p$ as $p=2q=\dfrac{2}{3}$ . Now we will find the required values as per given in the question one by one.
(i) When we have the value of $x=0$
So, the probability distribution can be found out by using the formula, $P\left( x \right)=n{{C}_{x}}{{p}^{x}}{{q}^{n-x}}$ . Here, we have $x=0,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3}$ . Substituting these values in the formula, we get,
$\begin{aligned} & P\left( 0 \right)=2{{C}_{0}}{{\left( \dfrac{2}{3} \right)}^{0}}{{\left( \dfrac{1}{3} \right)}^{2-0}} \\\ & \Rightarrow P\left( 0 \right)=2{{C}_{0}}{{\left( \dfrac{2}{3} \right)}^{0}}{{\left( \dfrac{1}{3} \right)}^{2}} \\\ & \Rightarrow P\left( 0 \right)=\dfrac{2!}{2!0!}\times \left( 1 \right)\times \left( \dfrac{1}{9} \right) \\\ & \Rightarrow P\left( 0 \right)=1\times 1\times \dfrac{1}{9} \\\ & \Rightarrow P\left( 0 \right)=\dfrac{1}{9} \\\ \end{aligned}$
(ii) When we have the value of $x=1$
So, here we will have the values as, $x=1,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3}$ . So, applying these in the formula, we will get,
$\begin{aligned} & P\left( 1 \right)=2{{C}_{1}}{{\left( \dfrac{2}{3} \right)}^{1}}{{\left( \dfrac{1}{3} \right)}^{2-1}} \\\ & \Rightarrow P\left( 1 \right)=2{{C}_{1}}{{\left( \dfrac{2}{3} \right)}^{1}}{{\left( \dfrac{1}{3} \right)}^{1}} \\\ & \Rightarrow P\left( 1 \right)=\dfrac{2!}{1!1!}\times \dfrac{2}{3}\times \dfrac{1}{3} \\\ & \Rightarrow P\left( 1 \right)=\dfrac{2\times 2}{9} \\\ & \Rightarrow P\left( 1 \right)=\dfrac{4}{9} \\\ \end{aligned}$
(iii) When we have the value of $x=2$
Now, we have the values as, $x=2,n=2,p=\dfrac{2}{3},q=\dfrac{1}{3}$ , so we get,
$\begin{aligned} & P\left( 2 \right)=2{{C}_{2}}{{\left( \dfrac{2}{3} \right)}^{2}}{{\left( \dfrac{1}{3} \right)}^{2-2}} \\\ & \Rightarrow P\left( 2 \right)=2{{C}_{2}}{{\left( \dfrac{2}{3} \right)}^{2}}{{\left( \dfrac{1}{3} \right)}^{0}} \\\ & \Rightarrow P\left( 2 \right)=\dfrac{2!}{2!0!}\times \dfrac{4}{9}\times 1 \\\ & \Rightarrow P\left( 2 \right)=\dfrac{4}{9} \\\ & \Rightarrow P\left( 1 \right)=\dfrac{4}{9} \\\ \end{aligned}$
Thus, we get the probability distribution of the number of heads as,

$x$	0	1	2
$P\left( x \right)$	$\dfrac{1}{9}$	$\dfrac{4}{9}$	$\dfrac{4}{9}$

So, we get the values of a = 1, b = 4, c = 4, and d = 9
So, the correct answer is “Option B”.

Note : We must note that if the coin was unbiased, then we would have taken the probability of getting head, $p=\dfrac{1}{2}$ and the probability of getting a tail, $q=\dfrac{1}{2}$ and the total probability would be equal to 1. We should also remember that binomial distribution can be considered only if there are two events, one is a success and the other is a failure as it was the case with the given question.